The basic circuit is at once simple and
super-effective. Film and low-ESR decoupling can
be used with it and it will not oscillate. Is there another circuit
that
is as effective with a comparable parts count?

Last updated
April 22, 2014Please help fund my R&D:

In
professional circles, designing analog circuits for good supply noise
tolerance is the accepted norm. But home experimenters have found
that low-noise power supplies present the ultimate freedom to test out
their inexhaustible creativity. The extra degree of freedom gained from
clean power allows for less distractions from the ultimate goal of a
good-sounding circuit. So audio hobbyists have been hard at work for
decades refining the art of quiet, inert power supplies.
Along these lines several super-high performance solutions are
available from the DIY community. Many of them are very complex,
needing performance opamps and many discrete devices. But as is often
the case, increasing complexity tends to give diminishing returns. For
this reason the simplest solutions are popular to hobbyists. My circuit
is a bit simpler than most, but I see few competitors where parts
count, cost and ease of assembly are concerned.

I'll use the positive
circuit to explain. Q1 and Q2 can be recognized by most poweramp
designers as a Complementary Follower Pair, CFP for short. This
arrangement provides dramatically reduced output impedance and
equivalent current gain to a Darlington arrangement - just what we want
for a power source. Both transistors need at least 1.2V Vce to work
best, which is what the triple diode string D1 is for. Q2 needs more
than Q1, so I've chosen 3 diodes for 1.2V. This may seem wrong to you,
but consider that the only current through these diodes is the base
current of Q1. Diode Vfb is roughly 400mV in the uA range, following
the 60mV/decade rule. Q1 is biased at about 15mA, which generally gives
the lowest output impedance. D2 and R8 determine the turn-on time, and
limit how much current the circuit will dump into proceeding supply
capacitors. This was important for one builder who employed a morbidly
gratuitous reservoir array. The first time he flipped the power switch
the transistors would blow from the inrush current! It took a few tries
for me to realize the problem... If your circuit does not have inrush
current over 1A, and this can be verified in simulation, you do not
need R8 and can bypass it. D3 discharges C1 when power is lost so it
doesn't discharge through Q1 - do not omit! The BC337/327 pair are crucial to this design - they set
the upper limit on input rejection. I got input rejection using these
of about 66db for the positive circuit and 54db for the negative
circuit (measured in real life). BC550C/560C are second-best in this
regard. ALL high-voltage transistors I have seen have bad
quasi-saturation behavior and should be avoided for low-Vce
applications (2N5551, 2SC1845, etc).

If we say that the
output impedance of the Kmultiplier is equivalent to that of
electrolytics, we establish a baseline to consider its advantages. One
incontrovertible advantage over electrolytics is that the Kmultiplier
has low impedance into the subsonic range. A comparable lytic in this
regard would be huge, and would necessitate soft-start circuitry to
prevent the massive inrush from destroying the preceeding wiring.
Clearly the Kmultiplier provides an expediency that no capacitor can
replace.

The
average electrolytic 470uF and up has lower than 50mR ESR. I have
measured this figure repeatedly on lytics pulled out of broken modern
entertainment devices such as computer monitors, TVs, and power amps.
The best have less than 20mR. So the output impedance of this very
simple circuit fares very well in practical terms. The fact that this
circuit has output impedance comparable to modern lytics AND will
tolerate film and low-ESR bypass shows that I have struck a good
balance - output impedance is not any lower than is actually needed,
but is low enough to augment and seamlessly integrate with a good
bypassing and decoupling scheme.

The voltage overhead of the Kmultiplier is low, making it a persuasive
drop-in addition to circuits which don't need DC regulated rails but
would benefit from heavy power filtering. In this case the Kmultiplier
replaces an oversized transformer and banks of capacitors, making it a
truly economical and expedient solution.

Because the Kmultiplier has no more feedback than needed and is
self-limiting in bandwidth, the compensation capacitors are merely
those parasitically included with the transistor junctions. This meager
amount of capacitive stress means that the circuit is unlikely to
glitch even with very fast load transients. Many regulators will emit a
highly distorted output signal when stressed with fast load current
swings, which can get into circuitry and make things worse than if the
supply was not regulated. The Kmultiplier however remains faithful into
the hundreds of KHz. Of course, why even risk high-speed signals at the
regulator when you can apply liberal decoupling across the load without
fear of oscillation?

Design Considerations

Choosing Components

Capacitors: R1 and the ESR of C1 form a resistive divider that limits the PSRR from the outset. Luckily, a cap with 100mR ESR as C1 will reduce PSRR by 1/5 in the worst-case scenario. The average cap in the 1000u rage has >50mR ESR, and going below this gives diminishing returns, so most noble brands should be safe.

Resistors: Resistor type will not affect the performance of the Kmultiplier significantly, unless maybe they are inductive. Film resistors will be fine.

Dissipation and heatsinking

Due
to the exceedingly low Vce of Q2 it does not need a heatsink; it will
hit Icmax before dissipation reaches 2W.

Controlling
inrush current

An
inrush of over 1A can destroy Q2, and often the LED as well. The RC
filter gives us a convenient and simple way to manage inrush. At
turn-on the full supply voltage minus the LED and diode string voltage
is imposed across R8. Thus if you have a 30V input, R8 will see about
25.2V. This gives C1 a max charging current of 25.2V/680R or 37mA.
Because the output follows the RC voltage, this means that for every
820uF of capacitance at the output, there will be 37mA of inrush. So
say you have 4700uF at the output. (4700uF/820uF)*37mA gives us about
212mA.

For
the circuit above, use this equation to find out what value of R8 you
will need to stay below 1A: (Vin-4.8)/(1A/(Cout/C1)) where Cout is your
TOTAL output capacitance.

Output impedance depends on output
current

This regulator exhibits diodic output impedance. This means
output impedance decreases with increased loading. The opposite is also
true, which is why this circuit should NEVER be used with loads less
then 25mA.

It cannot absorb negative load
surges

The transistors only conduct
in one direction - make sure your circuit will not generate negative
current surges.

Keep layout compact

This is a very fast circuit. It has behaved well for me so
far,
but long traces at sensitive nodes may encourage it to do things it
ordinarily wouldn't.

It is not a fixed-voltage
regulator - the output just lags the input by several seconds

The output voltage will track the input voltage over time -
this can be to your advantage. If your rails can vary by several volts,
you won't need the full voltage headroom you would need for a
fixed-voltage regulator. The Kmultiplier will always "relax" to 1.8V
voltage drop between input and output. However if your circuit needs a
precise and regulated supply voltage, this circuit may not be for you.

Troubleshooting

How do I know when it is working
correctly?

A very fine balance is required to maintain the high input
isolation. Therefore if anything is wrong, filtering will be impaired.
To verify that you Kmultiplier is working right, follow this procedure:

Turn on your Kmultiplier and have an AC voltmeter at hand.
Short the probes to make sure it can read below 1mVAC.

Measure and record the input AC voltage and the output AC
voltage.

Divide the measured output by the input. Make sure this
matches the specified input isolation.

The LED should flash when you flip the power switch.

If the LED is emitting ANY light at all during operation, it
means your
Kmultiplier is struggling and failing to meet the current demand, or
that there is a fault.

The voltage drop roughly corresponds to 3 diode junctions.
Measure the DC voltage between the input and output of your
Kmultiplier. It should be within .15V of the specified voltage drop.

Isolating the fault

Voltage drop too high

Measure the voltage across R2. If it is over 80mV, you're
drawing too much current or Q1 is faulty possibly from overcurrent or a
backwards discharge through the BE junction. C1 could also be faulty
due to aging or exposure to heat over time.

Voltage drop too low

If voltage drop is less than a diode (.6V) Q2 is faulty
from overcurrent, possibly inrush. Rarely, if that is not the problem,
D4 may be shorted.

LED does not flash during turn-on

Check your input voltage. Most likely a failure in
another part of the circuit has blown the LED. This will not change the
troubleshooting process.

If you want to change
something, beware that it may change troubleshooting procedures and
design considerations. That said, if you are confident you do not need
guidance for your application, here are some ideas and helpful
information. Also, discuss your modifications with us in the Kmultiplier
thread.

Decreasing voltage drop

One reason people use C-multipliers is because they can
tolerate low input voltages. You can decrease the voltage drop of the
Kmultiplier by taking diodes out of the diode string D1. Beware however
each time you do this you decrease the ripple tolerance by 800mV pk-pk.

Input ripple considerations

If you are increasing voltage drop in order to have more
ripple tolerance, pay attention to the LED breakover voltage. If you
expect to have more ripple than the specified tolerance for the
original circuit, then upgrade to a green LED instead, or a string of
red LEDs. Otherwise ripple will intermittently turn on the LED and ruin
input isolation.

The RC time constant will
determine how closely it tracks the input voltage

For better filtering you may try to increase the RC time
constant, but beware that the "stiffer" the filter, the more likely
that rail sag and power demands will saturate it and possibly cause
glitches.

This version is good for the frontend of very high-power amps with 100V rails, or any application where you need more than 45V output. It can be used at any rail voltage, regardless of the
transistors' individual voltage limits, as long as the capacitors are up to it. The only difference is the startup is immediate rather than delayed, and it needs a lot of diodes.

This version is for tube preamps. R5 limits output current but is necessary to ensure safe inrush.

Is there a Kmultiplier I can use for a
power amp?

Adapting
the Kmultiplier for higher currents is not so simple. An increased
number of active devices is required, and new feedback loops must be
added. This creates a pandora's box of new difficulties which are not
in the original design simply because of its simplicity. A more
powerful version needs compensations and very careful dimensioning to
achieve the benefits of low voltage drop, low output impedance and
compatibility with well-decoupled designs. The resulting circuit is not
so elegant in its presentation or its performance, but I think it can
be done, and I may have done it. I will link to it tentatively.

Theory of Operation

The design of this
circuit begun when I was thinking about the capacitance multiplier
circuit block, shown below. The C-multiplier is such a useful,
underused circuit. It can replace banks of caps, given enough voltage
headroom. Furthermore its input isolation can be a big bonus for RF
filtering provided you select transistors carefully. However it seemed
to me no one had thought about this circuit for decades. Relegated to
ancient history, no one seemed to grasp it's potential when revamped
with modern transistors and design expertise. Here is the whole story.

It is called the
capacitance multiplier because, (according to theory) it appears to the
load as if C1 has been multiplied by the Hfe of Q1. This particular
transistor has an Hfe of 150 or so, which means our load sees a supply
capacitance of about 1500uF.

Those who have gotten this far probably understand that the reality is
less pretty. After all, the BE junction of Q1 acts like a silicon diode
in series with the load. This means the output impedance of this filter
circuit is very high, and very nonlinear. The impedance of general
purpose silicon follows the rule R=.033/Id, where R is the small-signal
resistance of the junction and Id is it's forward bias current in the
given application. So if our load draws about 40mA, then our filter has
an output impedance of tentatively .825 ohms. This is abysmal.
As seen by the
load, R1 is also divided by the Hfe of Q1, and this is in series with
the "diodic" output impedance - 825mR+(100R/150Hfe)=1.492R. Even worse.
But this is followed by a
sigh of relief when we realize that at AC the transistor's base sees a
short through C1, returning the output impedance to 825mR.

Even so the
nearly doubled output resistance at DC looms over us.
Lets set
up a comparison to give a sense of reality and scale to these figures.
This circuit tries to emulate a 1500uF capacitor. A standard
electrolytic capacitor in this range has around 50mR or .05 ohms.
Furthermore, the transistor only conducts one way, unlike the capacitor
which can absorb negative current surges. So as you can see, the
circuit does not really compare to a real capacitor. They are so
different they cannot be treated like equivalents. The drawbacks are
listed as follows:

High output impedance

potentially much more than an
equivalent
electrolytic and even the rectifier and transformer winding.

That aside, the
circuit still has one considerable benefit over a bare
1500uF electrolytic. This is ripple rejection and input isolation. As
Q1 is in the emitter follower configuration, the emitter follows the
base voltage, and the base is fed by an RC lowpass filter of
100R*100uF. We don't get the benefit of this RC arrangement with
just a 1500uF capacitor - where the only R is that of the rectifier
diodes and the transformer winding. The input isolation is limited only
by the RC corner frequency and transistor leakage. Depending on the
transistor, you may expect AC input rejection from 40-70db. This is
where transistor selection makes all the difference. I won't go into
detail on all the ways to improve this dinosaur, but here are a number
of ways for you to consider if you don't really need the performance
that can be gained from an extra transistor:

Use a second-order RC filter

Much improves isolation near the
corner frequency, and at higher frequencies up to the leakage limit.

Use a diode string or LED in place of
the resistor

This will
make turn-on instant and dynamically adjust resistance to keep Vcb
equal to the diode breakover. But watch the input ripple and your
voltage drop.

Use an FET or MOSFET

This lets you make the resistor exceedingly large. What you
gain in corner frequency you may lose in leakage,
voltage drop, and output impedance. Perhaps this concept is more
applicable to active audio filters.

At this point there
are many, many things we could try to tailor the performance in many
directions. Of these, the options adding another active device tend to
become a bit less flexible. If we replace Q1 with a Darlington pair for
instance (diagram below), you will need to draw enough current at all
times to keep the driver transistor on. An extra diode drop is added,
but possibly offset by a lower R1 voltage drop. Even so, the
performance gains can be dramatic. Due to the greatly decreased base
current, we can increase R1 by several times. This allows output
filtering to the subsonics, or alternatively less resistive output
impedance at DC. However AC output impedance can be made worse. Say for
instance our driver transistor is biased at 2mA and our output
transistor has an Hfe of 100. The load draws 102mA. Following the
diodic output impedance rule, Q1 defines most of the output impedance
at 330mR. It's base current is 100mA/100, 1mA. The diodic resistance of
Q2's emitter is divided by the output's Hfe like in the original
C-multiplier, and comes to 165mR. So the total output impedance,
neglecting R1, is 495mR. Often times designers neglect to give the
driver any bias current at all except the base current of the
output. Because the output's base current rises proportionally with
load current, and the diodic emitter resistance decreases
proportionally with emitter current, the net result is that proportions
cancel, and the Darlington output resistance gains the nonlinearity of
2 diodes in series - 660mR. Neat, right? Of course, you had better
decouple the supplies well, because any fast load signals will saturate
and pump the driver transistor and result in nasty glitching.

Ultimately, the
Darlington C-multiplier still leaves us wanting more. Many designers
don't feel like pushing the limit - the C-multiplier was never
the Rolls Royce of supply solutions anyway. Why not just use an LM7812?
But because of reasons mentioned at the beginning of this page, even
that is not a satisfying option. Is there a middle ground between
simplicity, expediency and performance?

Most designers already know about the benefits of a CFP over a
Darlington, even though few seem to have thought to apply it to a
C-multiplier. The CFP has higher transconductance in common emitter
form, which translates to lower output impedance in common collector
form. However the general consensus on the CFP from amplifier designers
is that it is unstable and risky. Many early amplifier designs
featuring CFP output stages were found to up and blow up one day for no
apparent reason. Eventually it came out that the cause was that it
has a tendency to oscillate, causing the transistors to dissipate a lot
more power than they should have. For well-trained amplifier designers,
the problem is just a matter of engineering, but the circuit in
question must be measured in the prototype; RF parasitics depend so
heavily on wiring that it is easier just to probe with a signal
generator. Truth be told many designers don't have the
background necessary to understand the problem.

I can go into detail on what the problem is in another article, but
here I will discuss the principle of operation.

Here again is the Kmultiplier diagram for you to refer to throughout my
explanation:

Another name for the
CFP could be the "G-multiplier pair". In essence, the output
conductance
is the conductance of Q1 multiplied by the current gain of the
Q2/R1
arrangement. Let's analyze the situation and get an idea of how this
works.

Q1 is biased
into it's most linear range by R1. So Ic(Q1) is roughly .68/47=~15mA.
This is a bit low to accommodate for a nominal max of 5mA Ib(Q2). In
the nominal range of loading Ic(Q1) ranges from 15mA to 20mA. So, once
more following the diode rules, the output resistance of Q1 varies
between 2.2R and 1.65R. Now lets include Q2. Lets say our load current
is 116mA. This sets the Ic of Q1 and Q2 to 100mA and 16mA respectively,
accounting for Q2's Hfe of about 100. At 100mA, Q2's Re is about 330mR.
For every 1mA of loading, Vbe(Q2) increases 330uV. This increase in
voltage across R1 results in a 7uA increase in it's current. This is
added to the increase in Ib(Q2) of 10uA and we get an increase in
Ic(Q1) of 17uA. 17uA across Q2's Re of ~2.2R gives us a final 37uV
output drop per 1mA. 37uV/1mA gives us 37mR as the entire arrangement's
output impedance. This is close enough for horseshoes to the measured
values.